# Magnet falling though copper pipe

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• Orthoceras
In summary, a falling neodymium magnet is decelerated in a vertical copper pipe due to the currents produced by Lenz's Law, which generate an upward force that counteracts the force of gravity. The electrons participating in the current do not "fall down" the pipe, and if the copper pipe had no resistance and was infinitely long, the electrons would fall down in a frictionless medium. The terminal speed of the magnet can be calculated using a formula that takes into account variables such as the resistivity of copper, the strength of the magnetic field, and the mass of the magnet. The electrons are held up by electrostatic forces, and even without the presence of the magnet, they contribute to the rigidity of the metal due
Orthoceras
I do not fully understand why a falling neodymium magnet is decelerated in a vertical copper pipe. It is usually explained by Lenz's law, two induced currents generate an upward force that counteracts the force of gravity. (link) However, I assume this upward force has to be generated by the free electrons participating in the current, and I would think these electrons do not have a good foothold in the copper. Therefore, it seems more likely that the free electrons would fall downwards through the copper, together with the heavy magnet, instead of decelerating it.

Suppose the copper pipe was replaced by an infinitely long conductor having zero resistance, would that affect the terminal speed of the magnet?
Is there a formula for the terminal speed of the falling magnet in a copper pipe?

Orthoceras said:
I do not fully understand why a falling neodymium magnet is decelerated in a vertical copper pipe. It is usually explained by Lenz's law, two induced currents generate an upward force that counteracts the force of gravity. (link) However, I assume this upward force has to be generated by the free electrons participating in the current, and I would think these electrons do not have a good foothold in the copper. Therefore, it seems more likely that the free electrons would fall downwards through the copper, together with the heavy magnet, instead of decelerating it.

Suppose the copper pipe was replaced by an infinitely long conductor having zero resistance, would that affect the terminal speed of the magnet?
Is there a formula for the terminal speed of the falling magnet in a copper pipe?
The currents produced in the copper pipe (described by Lenz's Law) circulate around the pipe. The electrons do not "fall down" the pipe.
If the copper somehow had no resistance, it could levitate the magnet. Yes one can derive a formula. What do you think the salient variables are (that will be in the formula)?

/

vanhees71
According to Newton's third law, the electrons participating in the eddy current are inevitably pushed down. If the copper pipe has zero resistance, and is infinitely long, I would expect the electrons to fall down, in a frictionless medium.

If vertical motion of the electrons somehow would be prohibited, for example by replacing the vertical pipe by a stack of copper rings separated by gaps I would expect a terminal velocity to be proportional to the resistivity of copper, ρ. (Something like v ∝ ρBmg). But in the copper tube electrons are free to move vertically.

davenn
Orthoceras said:
these electrons do not have a good foothold in the copper.
At first, I thought this sentence sounded a bit naive. But once I got through the whole post and thought a bit more, I felt the question is actually not quite trivial.

If we mount the tube in some sort of force balance, then obviously some downward force will be registered as the magnet makes its way down. If the tube has some freedom to move, it will be dragged downwards. So we do need to ask what (upward) force acts on the electrons, such that they in turn can exert an upward force on the magnet.

I think it's just that the electrons do really begin to drift downwards a tiny distance, but then the new charge configuration immediately produces an electric potential that opposes any further drift. There may even be some clever way to measure the potential difference. Perhaps it may be valid to think of it as vaguely analogous to the Hall effect.

Edit: If we replace the tube with an elongated torus (in effect a double walled tube with the inner and outer walls shorted at the top and bottom), would the magnet drop faster, since the electrons are no longer impeded by a voltage buildup?

Edit^2 : If I remember correctly, the drift velocity of the electrons would be of the order of the speed of the magnet, considering the slowed-down motion of the latter. This suggests to me that the upward force on the electrons could just come from collisions with impurities, lattice defects etc --- aka plain old fashioned ohmic resistance. So in a nutshell, the magnet wants to drag the electrons downwards, and it does so -- only the electrons are heavily impeded by collisions/resistance which limits the drift rate, which in turn provides the upward force that we are looking for.

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The electrons are "held up" by electrostatic forces whether the magnet is present or not. If I put a piece of lead (same mass as Nd magnet) on the pipe it will compress slightly and the new electrostatic equilibrium will support the weight. Much th same thing with magnet but there are induced currents as well.

hutchphd said:
The electrons are "held up" by electrostatic forces whether the magnet is present or not. If I put a piece of lead (same mass as Nd magnet) on the pipe it will compress slightly and the new electrostatic equilibrium will support the weight. Much th same thing with magnet but there are induced currents as well.
This is clear enough as far as the bound electrons are concerned. But do the conduction / free / delocalized electrons also contribute to the rigidity of the metal? And do they experience the kind of forces you refer to?

Every electron interacts with every other charged particle in that chunk of metal (or nonmetal for that matter). If I leave a piece of pipe standing on the table do you think the "free" elctrons will puddle in the bottom of thepipe?
And the term "free" electrons needs to be understood in a very particular way. They are free and mobile because they are bound to a highly periodic scaffold which is the crystal lattice. Their wavelike nature affords them a freedom they would not otherwise possess.

Chally and DaveE
I think the current must be tilted down slightly due to the motion of the magnet. The magnet actually makes a lot of noise whilst descending, and I wondered if it actually rotates on its way down due to the current following a helical path?

If the pipe is vertical and the magnet aligned then by symmetry it will not rotate I believe. Which way would it go?

Another way to qualitatively understand the physics is as follows: A moving magnet also implies an electric field. The electric field sets the conduction electrons in the copper in motion leading to the eddy currents. Due to energy conservation this counteracts the free fall of the magnet through the force by the induced magnetic field of these currents.

ipsky and hutchphd
Orthoceras said:
Suppose the copper pipe was replaced by an infinitely long conductor having zero resistance, would that affect the terminal speed of the magnet?
Is there a formula for the terminal speed of the falling magnet in a copper pipe?
Computing terminal velocity is very difficult but if the pipe had zero resistance then the magnet would be indefinitely suspended in the pipe!

DaveE and hutchphd
rude man said:
if the pipe had zero resistance then the magnet would be indefinitely suspended in the pipe!
I am primarily interested in applying Newton's third law here. Suppose we connect both ends of the pipe by a copper wire having zero resistance. This eliminates the electrostatic force that previously kept the electrons suspended at a fixed height. In that case an indefinitely suspended magnet would be a violation of Newton's third law.

davenn
Orthoceras said:
In that case an indefinitely suspended magnet would be a violation of Newton's third law.
So then the magnet that's stuck to my refrigerator door is violating the 3rd law because it doesn't fall off?

Chally and hutchphd
Orthoceras said:
This eliminates the electrostatic force that previously kept the electrons suspended at a fixed height
No it will not. The metal also contains the exact same number of positive charges. What are you trying to get at ??

Ok, then I misunderstood why you introduced the electrostatic force in this discussion.

My point remains that the Lorentz force is pushing the electrons that are participating in the eddy current down, and there is no upward force exerted on them (if the resistance is zero, and both ends of the pipe are connected). So I am not sure that the magnet is indefinitely suspended if the resistance is zero.

davenn
Orthoceras said:
both ends of the pipe are connected

Or better still, (from my post above):
Swamp Thing said:
If we replace the tube with an elongated torus (in effect a double walled tube with the inner and outer walls shorted at the top and bottom), would the magnet drop faster?

If we don't neglect resistance, then the electrons' downward movement would be limited by the ohmic resistance (ultimately down to interaction with defects/impurities).

If we assume zero resistance, it's conceivable that the magnet would drag the eddying electrons down the inner wall, and current would then flow up the outer wall -- hence allowing the magnet to fall freely.

However, this closed loop will create a magnetic field in the volume between the inner and outer walls -- which could again, via Lenz's law, set up an impeding force.

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davenn
I will not try to point out all of the wrong physics being described in the last few posts.

If the copper pipe we replaced by an aligned stack of copper hoops insulated one from another, the physics would change almost not at all. The axial currents do not matter. The double walls essentially would not matter. The important induced currents are circumferential.

hutchphd said:
Yes one can derive a formula. What do you think the salient variables are (that will be in the formula)?
My assumption in this description is that the cylindrical magnet has a pole at each flat end and is a loose slip fit in the cylindrical pipe ..

davenn and Chally
hutchphd said:
I will not try to point out all of the wrong physics being described in the last few posts.
The physics could well be totally wrong, but the question that we're trying to answer is, I think, a valid one:

• The interaction between the falling magnet and the eddying electrons is such that it tends to drag them down (a la Newton's action-reaction)
• Something must keep them from drifting down.
• What is that something?
Is there a fallacy or incorrect assumption in the first bullet point, the second one, or in both?

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davenn
See #5.

hutchphd said:
See #5.
hutchphd said:
The electrons are "held up" by electrostatic forces

If i may, let me try to elaborate what I think you mean -- just to help pin down where I'm going wrong.

So: As soon as the magnet starts to drag the eddying electrons down, a charge imbalance begins to develop. We have a slight buildup of negative charge at the bottom and a corresponding positive charge at the top. This charge configuration produces it's own force on the eddying electrons, which is what holds them up.

Correct so far?

davenn
What force "drags the electrons down"? The induced electric force is circumferential. Do you understand Faraday,s Law?

Chally
Um, the magnet experiences an upward force from its interaction with the electrons, so the same electrons must experience an equal and opposite downwards force ?

If that's not correct, then that's where my misunderstanding arises.

OK once the charge is moving circumferentially what magnetic force is on it as the pole face gets near?

hutchphd said:
OK once the charge is moving circumferentially what magnetic force is on it as the pole face gets near?
Can we think of the circumferentially moving charge as a sheet solenoid, with it's (eg) North pole facing the magnet's North pole?

This is just speculation - so for the moment let's say I don't really know the answer to your question.

But is it not the case that we have to account for an equal and opposite upward force somehow? And have I understood your #5 correctly, which provides that accounting force?

Yes we do need to talk to Sir Isaac. But those conduction electrons are not really free. They are still strongly coupled to all the the rest of the metal and each other and most of the Newton forces go to the metal as a whole (even if they were" bouncing" end to end the net momentum ends up quickly in the entire pipe (i e the clamp holding it) Incidentally the "individual particle" picture of electrons in solids is not always great physics...
The residual B field will be radial and axial giving axial and radial IXB. Notice the effect of the trailing pole will do exactly the reverse.
Mostly a wave of circulating currents will accompany the magnet...not the electrons per se.

More than you want to know

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davenn, Chally, Einstein44 and 2 others
Orthoceras said:
I am primarily interested in applying Newton's third law here. Suppose we connect both ends of the pipe by a copper wire having zero resistance. This eliminates the electrostatic force that previously kept the electrons suspended at a fixed height. In that case an indefinitely suspended magnet would be a violation of Newton's third law.
No it would not.
1. There are no electrostatic forces involved here. The Lorentz force is electrodynamic, not electrostatic. There are no charges setting up any electric fields. The charges moving in the pipe are emf-induced.
2. You seem to think of gravity having any effect on the charges. It does not, in any practical sense.

hutchphd said:
What force "drags the electrons down"? The induced electric force is circumferential. Do you understand Faraday,s Law?

The Lorentz force on the electrons that participate in both eddy currents is downward, if the magnet is falling downward

hutchphd
Orthoceras said:
The Lorentz force on the electrons that participate in both eddy currents is downward, if the magnet is falling downward

View attachment 294892
It's not the Lorentz force. The Lorentz force is the force on a moving charge by a stationary B field.

In this case we use the Faraday law. That law states that an E field is induced in the metal circumferentially according to ## \oint \bf E \times \bf dl = - d\phi/dt. ## The minus sign is what makes your assumption as to axial force direction wrong.

If the calculus confuses you, use Lenz's law: the current buildup must be such as to counter the change in the incoming B flux. Looking at your bottom current, the B flux is increasing downward and the right-hand rule applied to the current pushes it upwards, opposite to gravity. Your picture is good, though, in every way including current directions.

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tech99
The Lorentz force is (in SI units)
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}).$$

hutchphd
vanhees71 said:
The Lorentz force is (in SI units)
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}).$$

Ah, then in #27 I meant only the magnetic part of the Lorentz force, $$\vec{F}= q (\vec{v} \times \vec{B})$$, where B is the magnetic field of the magnet. The magnetic part of the Lorentz force is downward, if the magnet is falling downward.

hutchphd
I think, the above quoted Youtube video, quoted in #25, is excellent. There it becomes entirely clear, how the "braking effect" comes about, and the student (!) uses the correct terminology, i.e., EMF instead of voltage etc. He also derives the non-relativistic equation of motion and solves them in a quantitative way, using the dipole approximation for the magnet and (tacitly) the quasistationary approximation of electrodynamics.

Einstein44, nsaspook, Lord Jestocost and 1 other person
An excellent video, except that it does not explicitly address the special case that I was asking about in my opening post, the case where both the pipe and the return wire (the wire connecting top and bottom of the pipe) have zero resistance. ("Does zero resistance affect the terminal speed of the magnet?") In the case of zero resistance, the average vertical speed of the electrons in the pipe is arbitrary, while the copper atoms are stationary. Then the velocity v from the formula F = k v, in the video at t=21:10, is the difference between the vertical velocity of the magnet and the vertical velocity of the electrons (not the copper atoms). This allows the magnet to fall down to the bottom of the pipe, its lowest energy level.

There was no mention of a return wire in your original post.
I am finished here. Good luck

Orthoceras said:
Ah, then in #27 I meant only the magnetic part of the Lorentz force, $$\vec{F}= q (\vec{v} \times \vec{B})$$, where B is the magnetic field of the magnet.
Yes, that's what I meant too. But the ## q \bf E ## part is irrelevant here too, since that component is directed along the E fields you showed in your diagram, not axially. It's what makes the electrons flow circumferentially in the pipe.
Orthoceras said:
The magnetic part of the Lorentz force is downward, if the magnet is falling downward.
Unfortunately, still wrong as I tried to explain.

vanhees71 said:
The Lorentz force is (in SI units)
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}).$$
If I'm not mistaken, the OP is referring to the fact that ##\vec{v} ## is the resultant of the magnet's velocity in the lab frame and the electrons' velocity within the metal due to the current flow induced by the magnet. Or, to make things a bit simpler, it may help to consider that the magnet is stationary and the tube is moving -- in which case ##\vec{v} ## is the resultant of the tube's velocity (which is shared by the electrons) and the velocity of the electrons within the tube, which is tangential to the circumference.

This leads to a question that naturally arises for a naive but interested non-expert like myself:
"If a moving magnet causes an electron to acquire a certain additional relative velocity component due to Lorenz force, then does that newly acquired relative velocity component result in a new extra component of the Lorenz force? Or is everything factored in already?"

To me it seems that this question could possibly underlie the OP's problem and its resolution. If the answer to the above question is "no, there is no new component of Lorenz force", then we need look no further, and this discussion could be closed. If the answer is "yes, there is one", then we still need to explain what keeps the electrons from moving under that extra component of Lorenz force, because in our case that component would be along the tube's axis. (the ##F_{Lor}## in post #27) .Such an explanation may possibly be found in post #25, first paragraph... I don't understand it entirely, but that could be just me.

As for the video, it is excellent, but it seems to presuppose that the tube can be considered as a series of ring-shaped slices, and that the electrons remain within their respective original slices during the whole process. This assumption may well be perfectly valid, but the question of "why" is, I think, not addressed? Not addressed either is the question of, what would happen if we provided the right conditions for the electrons to move downwards under the extra ##F_{Lor}## component, i.e. travel down the tube from slice to slice? Would these conditions not be facilitated if we provided an external path for charges to return from the bottom to the top?

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